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Math Help - solve for x

  1. #1
    Senior Member pacman's Avatar
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    solve for x

    Solve for x: x^4 + x^3 + x^2 + x + 1 = 0.
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  2. #2
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    If you draw the graph of x^4 + x^3 + x^2 + x + 1, it won't cut the x-axis.

    So, x^4 + x^3 + x^2 + x + 1 = 0 has no real roots
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  3. #3
    Senior Member pacman's Avatar
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    Thanks songoku, I have an UGLY solution for this, but i am hoping THAT someone may be able to solve this elegantly.
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  4. #4
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    If anyone's curious, the four complex roots are the following:
    x ≈ 0.309 + 0.951i
    x ≈ 0.309 - 0.951i
    x ≈ -0.809 + 0.588i
    x ≈ -0.809 - 0.588i
    (using a computer, of course )

    Polar form:
    x = \cos \frac{2\pi}{5} + i\sin \frac{2\pi}{5}
    x = \cos \frac{8\pi}{5} + i\sin \frac{8\pi}{5}
    x = \cos \frac{4\pi}{5} + i\sin \frac{4\pi}{5}
    x = \cos \frac{6\pi}{5} + i\sin \frac{6\pi}{5}


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  5. #5
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by pacman View Post
    Solve for x: x^4 + x^3 + x^2 + x + 1 = 0.
    Observe that (x-1)(x^4+x^3+x^2+x+1)=x^5-1. Hence the solution you want is the set of the four complex 5th roots of unity. No need to plot graph or use computer.
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  6. #6
    Senior Member pacman's Avatar
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    Hint: divide BS x^2, we have

    x^2 + x + 1 + 1/x + 1/x^2 = 0, rearranging it and then grouping it

    (x + 1/x)^2 + (x + 1/x) + 1 = 0

    It is now quadratic in x + 1/x, the result is quite ugly but it can generate a result . . .
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  7. #7
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by pacman View Post
    Hint: divide BS x^2, we have

    x^2 + x + 1 + 1/x + 1/x^2 = 0, rearranging it and then grouping it

    (x + 1/x)^2 + (x + 1/x) + 1 = 0

    It is now quadratic in x + 1/x, the result is quite ugly but it can generate a result . . .
    x+\frac{1}{x}=\frac{-1\pm\sqrt{-3}}{2}

    Then

    x=\frac{-\left(\frac{-1\pm\sqrt{-3}}{2}\right)\pm\sqrt{\left(\frac{-1\pm\sqrt{-3}}{2}\right)^2-4}}{2}

    Is this what you are saying?
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  8. #8
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    I think there's a typo in pacman's last post. Instead of
    \left(x + \frac{1}{x}\right)^2 + \left(x + \frac{1}{x}\right) + 1 = 0
    it should have been
    \left(x + \frac{1}{x}\right)^2 + \left(x + \frac{1}{x}\right) {\color{red}-}\; 1 = 0


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  9. #9
    Senior Member pacman's Avatar
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    i mixed it up in a bowl of soup, yeongil is correct with his observation . . . .
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